Question

Light of wavelength ${\lambda }_0$ in air enters a medium of refractive index $n$. If two points $A$ and $B$ in this medium lid along the path of this light at a distance $x$, then phase difference ${\varphi }_0$ between these two points is

• A. ${\varphi }_0=\frac{1}{n}\left(\frac{2\pi }{{\lambda }_0}\right)x$
• B. ${\varphi }_0=n\left(\frac{2\pi }{{\lambda }_0}\right)x$
• C. ${\varphi }_0=(n-1)\left(\frac{2\pi }{{\lambda }_0}\right)x$
• D. ${\varphi }_0=\frac{1}{(n-1)}\left(\frac{2\pi }{{\lambda }_0}\right)x$

Answer 1

The correct option is B ${\varphi }_0=n\left(\frac{2\pi }{{\lambda }_0}\right)x$

${\lambda }_o$ is the wavelength of light in air.

Let $t$ be the distance between points S and A.

Optical path length travelled by light via SB $=n(t+x)$

Optical path length travelled by light via SA $=nt$

Path difference between points A and B $\Delta x=SB-SA=nx$

Now ${\varphi }_o=\Delta x\times \frac{2\pi }{{\lambda }_o}$

Hence, ${\varphi }_o=n\left(\frac{2\pi }{{\lambda }_o}\right)x$